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Electric field problem

  1. Sep 4, 2006 #1
    Three point charges are placed on the y-axis. a chage at q at y=a, a charge at -2q at the origin, and a charge q at y=-a. find the magnitude and direction of the electrif field on the y-axis for y>a

    I'm not sure where to begin at all. This is what I am thinking, but not completely sure. I could find the electric field the charges but I dont know how to encoporate the fact that there are two other charges that add to the electric field. any suggestions?
     
  2. jcsd
  3. Sep 4, 2006 #2

    Office_Shredder

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    You can calculate the electric field for each charge individually at a point, then add the three fields together (keeping in mind that opposite direction fields subtract, not add).
     
  4. Sep 5, 2006 #3
    k=1/4piε


    [tex]E=k \frac{-q}{4a^2} + k \frac{-2q}{a^2} [/tex]

    [tex] E= k \left( \frac{-q}{4a^2} + k \frac{-8q}{4a^2} \right)[/tex]

    [tex]E=k \left( \frac{-9q}{4a^2} \right) [/tex]

    does this look right?
     
  5. Sep 5, 2006 #4
    you aren't taking into account the differences in distances to each charge. Lets say you are at a point b along the y axis. then the distance 'r' from the first charge q is b-a. the distance for the one at the orgin (the -2q) is b-0 or b. lastly the distance for the one at -a is b+a. these are the radii you need to plug into the general equation
    E=kq/r^2.
     
  6. Sep 5, 2006 #5
    the -2q distance is a

    the -q distance is 2a

    correct?

    that is what I have.
     
  7. Sep 5, 2006 #6
    the problem stated says 'find the magnitude and direction of the electrif field on the y-axis for y>a'. Therefore the distance is not a set value, but depends on where you want to look (for values greater than a). This is why I used b. b is just an arbitrary y value.
     
  8. Sep 5, 2006 #7
    then the solution can be infinite!?

    actually, there is a second part to the problem. it asks me to use the binomial expansion to show that very far from the quadrupole the electric field is proportional to y^-4 which i, again, have no clue how to tackle.
     
  9. Sep 5, 2006 #8
    well the solution won't be infinite. as the radius goes to infinity the solution goes to zero since the radius is on the bottom.

    also not sure about the second part... i might try setting y>>a so that the radii all just go towards y... but that doesn't give you y^-4
    or try and use the taylor expansion and throw out small terms...
    taylor expansion
    (1+x)^r = 1 + rx + 1/2*r(r-1)x^2+1/6*r(r-1)(r-2)x^3...etc
     
  10. Sep 5, 2006 #9
    i see what you are saying

    b>a

    [tex]E=k \frac{-q}{(a+b)^2} + k \frac{-2q}{b^2} [/tex]
     
  11. Sep 5, 2006 #10
    close, but unless I am reading the problem wrong there are 3 charges. And two are positive q (with radii a-b and a+b). The last term is correct though.
     
  12. Sep 5, 2006 #11
    [tex]E=k \frac{q}{(b-a)^2} + k \frac{-2q}{b^2} + k \frac{q}{(a+b)^2} [/tex]
     
  13. Sep 5, 2006 #12
    yes, looks good
     
  14. Sep 6, 2006 #13

    how do you tell if it goes towards y^-4 or not?
     
  15. Sep 6, 2006 #14
    my guess would be to attempt to combine all of the fractions into one and then go from there.
     
  16. Sep 6, 2006 #15

    HallsofIvy

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    Your original problem was "find the magnitude and direction of the electrif field on the y-axis for y>a". None of your formulas include "y"!

    From the point at (0,y) to each of the charges, at (0, a), (0, 0), and (0, -a) the distances are |y-a|, |y|, and |y+a| respectively.
     
  17. Sep 6, 2006 #16
    thats my bad, i had started using b... b is y.
     
  18. Sep 6, 2006 #17
    the next part of the question asks: Use the binomial expansion to show that very far from the quadrupole, so that y>>a, the electric field is proportional to y^-4. Contrast the behavior to that of the eectric field of a point charge and that of the electric field of an electric dipole.

    [tex]E=k \frac{q}{(y-a)^2} + k \frac{-2q}{y^2} + k \frac{q}{(a+y)^2} [/tex]


    when y>>a

    [tex]E=k \frac{q}{y^2} + k \frac{-2q}{y^2} + k \frac{q}{y^2} [/tex]


    so E=0

    not really sure where to go from here
     
  19. Sep 6, 2006 #18
    before assuming that y>>a try moving the fractions into one.
    for example, the first term would become:
    [tex] k\frac{q y^2 (a+y)^2}{(y-a)^2y^2(y+a)^2} [/tex]
    similarly try this with the other two fractions and then they can be added into one.
     
  20. Sep 7, 2006 #19
    [tex]E= kq \left( \frac{ y^2 (a+y)^2}{(y-a)^2y^2(y+a)^2} - \frac{2 (y-a)^2 (a+y)^2}{(y-a)^2y^2(y+a)^2}+ \frac{ y^2 (a-y)^2}{(y-a)^2y^2(y+a)^2} \right)[/tex]

    assuming that y>>a

    [tex]E=kq \frac{y^4-y^4+y^4}{y^6}[/tex]

    I get the answer as y^-2

    where am I going wrong? and why does the question as me to use the binomial expansion?
     
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